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An absolutely continuous function on a compact interval is differentiable almost everywhere on that interval.

A measure $\mu$ is absolutely continuous with respect to another measure $\nu$ (denoted $\mu \ll \nu$) if $\nu(E) = 0$ implies $\mu(E) = 0$ for every measurable set E.

A function f is absolutely continuous on an interval [a, b] if for every $\epsilon > 0$ there exists $\delta > 0$ such that for any finite collection of disjoint sub-intervals ${[x_k, y_k]}$, the sum of the lengths of the sub-intervals is less than $\delta$ implies the sum of $|f(y_k) - f(x_k)|$ is less than $\epsilon$.

The total variation of an absolutely continuous function $f$ on an interval [a, b] can be computed as the integral of the absolute value of its derivative: $V_{a}^{b}(f) = \int_{a}^{b} |f'(x)| dx$.

Lebesgue's theorem states that a continuous function of bounded variation on [a, b] is absolutely continuous if and only if its set of discontinuities for its derivative is of Lebesgue measure zero.

The Radon-Nikodym theorem provides that if $\mu \ll \nu$, then there exists an integrable function $f$ such that $\mu(E) = \int_E f d\nu$ for all measurable sets E.

A function that is absolutely continuous on an interval is also uniformly continuous on that interval, but the converse is not necessarily true.

If a function is absolutely continuous on [a, b], then it is continuous on [a, b] and has a derivative $f'$ almost everywhere on [a, b], such that the integral of $f'$ over [a, b] equals $f(b) - f(a)$.

If $f$ is differentiable almost everywhere on [a, b], has a Lebesgue integrable derivative $f'$, and $f(x) = f(a) + \int_{a}^{x} f'(t) dt$ for all $x$ in [a, b], then $f$ is absolutely continuous on [a, b].